Thermodynamics · 18.1 The Three Laws of Thermodynamics In Chapter 6 we encountered the first of...
Transcript of Thermodynamics · 18.1 The Three Laws of Thermodynamics In Chapter 6 we encountered the first of...
The production of quicklime (CaO)from limestone (CaCO3) in a rotary kiln at about 9508C.
Thermodynamics
ESSENTIAL CONCEPTS
Laws of Thermodynamics The laws of thermodynamics havebeen successfully applied to the study of chemical and physicalprocesses. The first law of thermodynamics is based on the lawof conservation of energy. The second law of thermodynamicsdeals with natural or spontaneous processes. The function thatpredicts the spontaneity of a reaction is entropy. The second lawstates that for a spontaneous process, the change in the entropy ofthe universe must be positive. The third law enables us to deter-mine absolute entropy values.
Gibbs Free Energy Gibbs free energy helps us to determine thespontaneity of a reaction by focusing only on the system. Thechange in Gibbs free energy for a process is made up of twoterms: a change in enthalpy and a change in entropy timestemperature. At constant temperature and pressure, a decrease inGibbs free energy signals a spontaneous reaction. The change inthe standard Gibbs free energy can be related to the equilibriumconstant of a reaction.
Thermodynamics in Living Systems Many reactions of biologi-cal importance are nonspontaneous. By coupling such reactionsto those that have a negative Gibbs free energy change with theaid of enzymes, the net reaction can be made to proceed to yieldthe desired products.
C H A P T E R
Activity Summary
1. Interactivity: Entropies of Reactions (18.4)
2. Interactivity: Entropy vs. Temperature (18.4)
3. Interactivity: Free Energy—Equilibrium (18.6)
CHAPTER OUTLINE
18.1 The Three Laws of Thermodynamics 611
18.2 Spontaneous Processes 611
18.3 Entropy 612Microstates and Entropy • Changes in Entropy •Standard Entropy
18.4 The Second Law of Thermodynamics 617Entropy Changes in the System • Entropy Changes in theSurroundings • The Third Law of Thermodynamics and Absolute Entropy
18.5 Gibbs Free Energy 622Standard Free-Energy Change • Applications of Equation (18.10)
18.6 Free Energy and Chemical Equilibrium 629
18.7 Thermodynamics in Living Systems 632
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18.1 The Three Laws of Thermodynamics
In Chapter 6 we encountered the first of three laws of thermodynamics, which says
that energy can be converted from one form to another, but it cannot be created or
destroyed. One measure of these changes is the amount of heat given off or absorbed
by a system during a constant-pressure process, which chemists define as a change in
enthalpy (DH ).
The second law of thermodynamics explains why chemical processes tend to
favor one direction. The third law is an extension of the second law and will be exam-
ined briefly in Section 18.4.
18.2 Spontaneous Processes
One of the main objectives in studying thermodynamics, as far as chemists are con-
cerned, is to be able to predict whether or not a reaction will occur when reactants
are brought together under a specific set of conditions (for example, at a certain tem-
perature, pressure, and concentration). This knowledge is important whether one is
synthesizing compounds in a research laboratory, manufacturing chemicals on an
industrial scale, or trying to understand the intricate biological processes in a cell. A
reaction that does occur under the given set of conditions is called a spontaneous reac-
tion. If a reaction does not occur under specified conditions, it is said to be non-
spontaneous. We observe spontaneous physical and chemical processes every day,
including many of the following examples:
• A waterfall runs downhill, but never up, spontaneously.
• A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar
does not spontaneously reappear in its original form.
• Water freezes spontaneously below 08C, and ice melts spontaneously above 08C
(at 1 atm).
• Heat flows from a hotter object to a colder one, but the reverse never happens
spontaneously.
• The expansion of a gas into an evacuated bulb is a spontaneous process [Figure
18.1(a)]. The reverse process, that is, the gathering of all the molecules into one
bulb, is not spontaneous [Figure 18.1(b)].
18.2 Spontaneous Processes 611
A spontaneous reaction does not
necessarily mean an instantaneous
reaction.
A spontaneous and anonspontaneous process.© Harry Bliss, Originally published
in New Yorker Magazine.
Figure 18.1(a) A spontaneous process. (b) A nonspontaneous process.
(b)
(a)
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• A piece of sodium metal reacts violently with water to form sodium hydroxide
and hydrogen gas. However, hydrogen gas does not react with sodium hydroxide
to form water and sodium.
• Iron exposed to water and oxygen forms rust, but rust does not spontaneously
change back to iron.
These examples show that processes that occur spontaneously in one direction cannot,
under the same conditions, also take place spontaneously in the opposite direction.
If we assume that spontaneous processes occur so as to decrease the energy of a
system, we can explain why a ball rolls downhill and why springs in a clock unwind.
Similarly, a large number of exothermic reactions are spontaneous. An example is the
combustion of methane:
Another example is the acid-base neutralization reaction:
But consider a solid-to-liquid phase transition such as
In this case, the assumption that spontaneous processes always decrease a system’s
energy fails. Experience tells us that ice melts spontaneously above 08C even though
the process is endothermic. Another example that contradicts our assumption is the
dissolution of ammonium nitrate in water:
This process is spontaneous, and yet it is also endothermic. The decomposition of
mercury(II) oxide is an endothermic reaction that is nonspontaneous at room temper-
ature, but it becomes spontaneous when the temperature is raised:
From a study of the examples mentioned and many more cases, we come to the fol-
lowing conclusion: Exothermicity favors the spontaneity of a reaction but does not
guarantee it. Just as it is possible for an endothermic reaction to be spontaneous, it is
possible for an exothermic reaction to be nonspontaneous. In other words, we cannot
decide whether or not a chemical reaction will occur spontaneously solely on the basis
of energy changes in the system. To make this kind of prediction we need another
thermodynamic quantity, which turns out to be entropy.
18.3 Entropy
In order to predict the spontaneity of a process, we need to introduce a new thermo-
dynamic quantity called entropy. Entropy (S) is often described as a measure of how
spread out or dispersed the energy of a system is among the different possible ways
that system can contain energy. The greater the dispersal, the greater is the entropy.
Most processes are accompanied by a change in entropy. A cup of hot water has a
certain amount of entropy due to the dispersal of energy among the various energy
states of the water molecules (for example, energy states associated with the
¢H° 5 90.7 kJ/mol2HgO(s)¡ 2Hg(l ) 1 O2(g)
¢H° 5 25 kJ/molNH4NO3(s) ¡H2O
NH41(aq) 1 NO3
2(aq)
¢H° 5 6.01 kJ/molH2O(s)¡ H2O(l )
¢H° 5 256.2 kJ/molH1(aq) 1 OH2(aq)¡ H2O(l )
¢H° 5 2890.4 kJ/molCH4(g) 1 2O2(g)¡ CO2(g) 1 2H2O(l )
612 CHAPTER 18 Thermodynamics
Because of activation energy barrier, an
input of energy is needed to get this
reaction started.
When heated, HgO decomposesto give Hg and O2.
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translational, rotational, and vibrational motions of the water molecules). If left stand-
ing on a table, the water loses heat to the cooler surroundings. Consequently, there is
an increase in entropy because of the dispersal of energy over a great many energy
states of the air molecules.
As another example, consider the situation depicted in Figure 18.1. Before the valve
is opened, the system possesses a certain amount of entropy. Upon opening the valve,
the gas molecules now have access to the combined volume of both bulbs. A larger vol-
ume for movement results in a narrowing of the gap between translational energy lev-
els of the molecules. Consequently, the entropy of the system increases because closely
spaced energy levels leads to a greater dispersal among the energy levels.
Microstates and Entropy
Before we introduce the second law of thermodynamics, which relates entropy change
(increase) to spontaneous processes, it is useful to first provide a proper definition of
entropy. To do so let us consider a simple system of four molecules distributed
between two equal compartments, as shown in Figure 18.2. There is only one way to
arrange all the molecules in the left compartment, four ways to have three molecules
in the left compartment and one in the right compartment, and six ways to have two
molecules in each of the two compartments. The eleven possible ways of distributing
the molecules are called microscopic states or microstates and each set of similar
microstates is called a distribution.† As you can see, distribution III is the most prob-
able because there are six microstates or six ways to achieve it and distribution I is
the least probable because it has one microstate and therefore there is only one way
to achieve it. Based on this analysis, we conclude that the probability of occurrence
18.3 Entropy 613
†Actually there are still other possible ways to distribute the four molecules between the two compartments.We can have all four molecules in the right compartment (one way) and three molecules in the right com-partment and one molecule in the left compartment (four ways). However, the distributions shown in Figure18.2 are sufficient for our discussion.
Figure 18.2Some possible ways of distribut-ing four molecules between twoequal compartments. Distribu-tion I can be achieved in onlyone way (all four molecules inthe left compartment) and hasone microstate. Distribution IIcan be achieved in four waysand has four microstates. Dis-tribution III can be achieved insix ways and has sixmicrostates.
I
Distribution
III
II
Microstates
1
43
2
1
324
21
43
1
2 4
3
1
4 3
2
1
3 4
2
1
423
12
43
3
4 2
1
2
3 4
1
2
4 3
1
Translational motion is motion through
space of the whole molecule.
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of a particular distribution (state) depends on the number of ways (microstates) in
which the distribution can be achieved. As the number of molecules approaches
macroscopic scale, it is not difficult to see that they will be evenly distributed between
the two compartments because this distribution has many, many more microstates than
all other distributions.
In 1868 Boltzmann showed that the entropy of a system is related to the natural
log of the number of microstates (W ):
(18.1)
where k is called the Boltzmann constant (1.38 3 10223 J/K). Thus, the larger the W,
the greater is the entropy of the system. Like enthalpy, entropy is a state function (see
Section 6.3). Consider a certain process in a system. The entropy change for the
process, DS, is
(18.2)
where Si and Sf are the entropies of the system in the initial and final states, respec-
tively. From Equation (18.1) we can write
(18.3)
where Wi and Wf are the corresponding numbers of microstates in the initial and final
state. Thus, if Wf . Wi, DS . 0 and the entropy of the system increases.
Changes in Entropy
Earlier we described the increase in entropy of a system as a result of the increase in
the dispersal of energy. There is a connection between the qualitative description of
entropy in terms of dispersal of energy and the quantitative definition of entropy in
terms of microstates given by Equation (18.1). We conclude that
• A system with fewer microstates (smaller W ) among which to spread its energy
(small dispersal) has a lower entropy.
• A system with more microstates (larger W ) among which to spread its energy
(large dispersal) has a higher entropy.
Next, we will study several processes that lead to a change in entropy of a system in
terms of the change in the number of microstates of the system.
Consider the situations shown in Figure 18.3. In a solid the atoms or molecules
are confined to fixed positions and the number of microstates is small. Upon melt-
ing, these atoms or molecules can occupy many more positions as they move away
from the lattice points. Consequently, the number of microstates increases because
there are now many more ways to arrange the particles. Therefore, we predict this
“order disorder” phase transition to result in an increase in entropy because the
number of microstates has increased. Similarly, we predict the vaporization process
will also lead to an increase in the entropy of the system. The increase will be con-
siderably greater than that for melting, however, because molecules in the gas phase
occupy much more space, and therefore there are far more microstates than in the
liquid phase. The solution process usually leads to an increase in entropy. When a
sugar crystal dissolves in water, the highly ordered structure of the solid and part of
¡
5 k ln Wf
Wi
¢S 5 k ln Wf 2 k ln Wi
¢S 5 Sf 2 Si
S 5 k ln W
614 CHAPTER 18 Thermodynamics
Engraved on Ludwig Boltzmann’stombstone in Vienna is his famousequation. The “log” stands for“loge,” which is the naturallogarithm or ln.
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the ordered structure of water break down. Thus, the solution has a greater number
of microstates than the pure solute and pure solvent combined. When an ionic solid
such as NaCl dissolves in water, there are two contributions to entropy increase: the
solution process (mixing of solute with solvent) and the dissociation of the com-
pound into ions:
More particles lead to a greater number of microstates. However, we must also con-
sider hydration, which causes water molecules to become more ordered around the
ions. This process decreases entropy because it reduces the number of microstates of
the solvent molecules. For small, highly charged ions such as Al31 and Fe31, the
decrease in entropy due to hydration can outweigh the increase in entropy due to mix-
ing and dissociation so that the entropy change for the overall process can actually
be negative. Heating also increases the entropy of a system. In addition to transla-
tional motion, molecules can also execute rotational motions and vibrational motions
(Figure 18.4). As the temperature is increased, the energies associated with all types
of molecular motion increase. This increase in energy is distributed or dispersed
among the quantized energy levels. Consequently, more microstates become available
at a higher temperature; therefore, the entropy of a system always increases with
increasing temperature.
Standard Entropy
Equation (18.1) provides a useful molecular interpretation of entropy, but is normally
not used to calculate the entropy of a system because it is difficult to determine the
NaCl(s) ¡H2O
Na1(aq) 1 Cl2(aq)
18.3 Entropy 615
Solid Liquid
Liquid Vapor
Solute Solution
System at T1 System at T2 (T2 > T1)
(a)
(b)
(c)
(d)
Solvent
Figure 18.3Processes that lead to anincrease in entropy of the sys-tem: (a) melting: Sliquid .
Ssolid; (b) vaporization: Svapor .
Sliquid; (c) dissolving; (d) heat-ing: .ST2
7 ST1
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number of microstates for a macroscopic system containing many molecules. Instead,
entropy is obtained by calorimetric methods. In fact, as we will see shortly, it is pos-
sible to determine the absolute value of entropy of a substance, called absolute
entropy, something we cannot do for energy or enthalpy. Standard entropy is the
absolute entropy of a substance at 1 atm and 258C. (Recall that the standard state
refers only to 1 atm. The reason for specifying 258C is that many processes are car-
ried out at room temperature.) Table 18.1 lists standard entropies of a few elements
and compounds; Appendix 2 provides a more extensive listing. The units of entropy
are J/K or J/K ? mol for 1 mole of the substance. We use joules rather than kilojoules
because entropy values are typically quite small. Entropies of elements and com-
pounds are all positive (that is, S8 . 0). By contrast, the standard enthalpy of for-
mation (DH8f) for elements in their stable form is arbitrarily set equal to zero, and for
compounds it may be positive or negative.
Referring to Table 18.1, we see that the standard entropy of water vapor is greater
than that of liquid water. Similarly, bromine vapor has a higher standard entropy than
liquid bromine, and iodine vapor has a greater standard entropy than solid iodine. For
different substances in the same phase, molecular complexity determines which ones
have higher entropies. Both diamond and graphite are solids, but diamond has a more
ordered structure and hence a smaller number of microstates (see Figure 12.22).
Therefore, diamond has a smaller standard entropy than graphite. Consider the natu-
ral gases methane and ethane. Ethane has a more complex structure and hence more
ways to execute molecular motions, which also increase its microstates. Therefore,
ethane has a greater standard entropy than methane. Both helium and neon are
monatomic gases, which cannot execute rotational or vibrational motions, but neon
has a greater standard entropy than helium because its molar mass is greater. Heav-
ier atoms have more closely spaced energy levels so there is a greater distribution of
the atoms’ energy among the energy levels. Consequently, there are more microstates
associated with these atoms.
616 CHAPTER 18 Thermodynamics
z
y
Stable form
Stretched
Compressed
(b)(a)
x
Figure 18.4(a) A diatomic molecule canrotate about the y- and z-axes(the x-axis is along the bond).(b) Vibrational motion of adiatomic molecule. Chemicalbonds can be stretched andcompressed like a spring.
S8
Substance (J/K ? mol)
H2O(l) 69.9
H2O(g) 188.7
Br2(l) 152.3
Br2(g) 245.3
I2(s) 116.7
I2(g) 260.6
C (diamond) 2.4
C (graphite) 5.69
CH4 (methane) 186.2
C2H6 (ethane) 229.5
He(g) 126.1
Ne(g) 146.2
TABLE 18.1
Standard Entropy
Values (S 8) for Some
Substances at 258C
Predict whether the entropy change is greater or less than zero for each of the following
processes: (a) freezing ethanol, (b) evaporating a beaker of liquid bromine at room
temperature, (c) dissolving glucose in water, (d) cooling nitrogen gas from 808C to
208C.
(Continued )
Example 18.1
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18.4 The Second Law of Thermodynamics 617
Bromine is a fuming liquid atroom temperature.
Strategy To determine the entropy change in each case, we examine whether the
number of microstates of the system increases or decreases. The sign of DS will be
positive if there is an increase in the number of microstates and negative if the number
of microstates decreases.
Solution (a) Upon freezing, the ethanol molecules are held rigidly in position. This
phase transition reduces the number of microstates and therefore the entropy
decreases; that is, DS , 0.
(b) Evaporating bromine increases the number of microstates because the Br2 molecules
can occupy many more positions in nearly empty space. Therefore DS . 0.
(c) Glucose is a nonelectrolyte. The solution process leads to a greater dispersal of
matter due to the mixing of glucose and water molecules so we expect DS . 0.
(d) The cooling process decreases various molecular motions. This leads to a decrease
in microstates and so DS , 0.
Practice Exercise How does the entropy of a system change for each of the
following processes? (a) condensing water vapor, (b) forming sucrose crystals from a
supersaturated solution, (c) heating hydrogen gas from 608C to 808C, and (d) subliming
dry ice.
Similar problem: 18.5.
Just talking about entropy increases its
value in the universe.
18.4 The Second Law of Thermodynamics
The connection between entropy and the spontaneity of a reaction is expressed by the
second law of thermodynamics: The entropy of the universe increases in a sponta-
neous process and remains unchanged in an equilibrium process. Because the uni-
verse is made up of the system and the surroundings, the entropy change in the
universe (DSuniv) for any process is the sum of the entropy changes in the system
(DSsys) and in the surroundings (DSsurr). Mathematically, we can express the second
law of thermodynamics as follows:
For a spontaneous process: (18.4)
For an equilibrium process: (18.5)
For a spontaneous process, the second law says that DSuniv must be greater than zero,
but it does not place a restriction on either DSsys or DSsurr. Thus, it is possible for
either DSsys or DSsurr to be negative, as long as the sum of these two quantities is
greater than zero. For an equilibrium process, DSuniv is zero. In this case, DSsys and
DSsurr must be equal in magnitude, but opposite in sign. What if for some hypothet-
ical process we find that DSuniv is negative? What this means is that the process is
not spontaneous in the direction described. Rather, it is spontaneous in the opposite
direction.
Entropy Changes in the System
To calculate DSuniv, we need to know both DSsys and DSsurr. Let us focus first on DSsys.
Suppose that the system is represented by the following reaction:
aA 1 bB¡ cC 1 dD
¢Suniv 5 ¢Ssys 1 ¢Ssurr 5 0
¢Suniv 5 ¢Ssys 1 ¢Ssurr 7 0
Interactivity:Entropies of ReactionsARIS, Interactives
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As is the case for the enthalpy of a reaction [see Equation (6.17)], the standard
entropy of reaction DS 8rxn is given by the difference in standard entropies between
products and reactants:
(18.6)
or, in general, using S to represent summation and m and n for the stoichiometric
coefficients in the reaction,
(18.7)
The standard entropy values of a large number of compounds have been measured in
J/K ? mol. To calculate DS8rxn (which is DSsys), we look up their values in Appendix
2 and proceed according to Example 18.2.
¢S°rxn 5 ©nS°(products) 2 ©mS°(reactants)
¢S°rxn 5 [cS°(C) 1 dS°(D)] 2 [aS°(A) 1 bS°(B)]
618 CHAPTER 18 Thermodynamics
Similar problems: 18.11 and 18.12.
From the standard entropy values in Appendix 2, calculate the standard entropy changes
for the following reactions at 258C.
(a)
(b)
(c)
Strategy To calculate the standard entropy of a reaction, we look up the standard
entropies of reactants and products in Appendix 2 and apply Equation (18.7). As in the
calculation of enthalpy of reaction [see Equation (6.18)], the stoichiometric coefficients
have no units, so DS 8rxn is expressed in units of J/K ? mol.
Solution(a)
Thus, when 1 mole of CaCO3 decomposes to form 1 mole of CaO and 1 mole of
gaseous CO2, there is an increase in entropy equal to 160.5 J/K ? mol.
(b)
This result shows that when 1 mole of gaseous nitrogen reacts with 3 moles of
gaseous hydrogen to form 2 moles of gaseous ammonia, there is a decrease in
entropy equal to 2199 J/K ? mol.
(c)
Thus, the formation of 2 moles of gaseous HCl from 1 mole of gaseous H2 and
1 mole of gaseous Cl2 results in a small increase in entropy equal to 20 J/K ? mol.
Comment The DS 8rxn values all apply to the system.
Practice Exercise Calculate the standard entropy change for the following reactions
at 258C:
(a)
(b)
(c) 2NaHCO3(s)¡ Na2CO3(s) 1 H2O(l ) 1 CO2(g)
3O2(g)¡ 2O3(g)
2CO(g) 1 O2(g)¡ 2CO2(g)
5 20 J/K # mol
5 (2)(187 J/K # mol) 2 [(131J/K # mol) 1 (223 J/K # mol)]
¢S°rxn 5 [2S°(HCl)] 2 [S°(H2) 1 S°(Cl2)]
5 2199 J/K # mol
5 (2)(193 J/K # mol) 2 [(192 J/K # mol) 1 (3)(131 J/K # mol)]
¢S°rxn 5 [2S°(NH3)] 2 [S°(N2) 1 3S°(H2)]
5 160.5 J/K # mol
5 [(39.8 J/K # mol) 1 (213.6 J/K # mol)] 2 (92.9 J/K # mol)
¢S°rxn 5 [S°(CaO) 1 S°(CO2)] 2 [S°(CaCO3)]
H2(g) 1 Cl2(g)¡ 2HCl(g)
N2(g) 1 3H2(g)¡ 2NH3(g)
CaCO3(s)¡ CaO(s) 1 CO2(g)
Example 18.2
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The results of Example 18.2 are consistent with those observed for many other
reactions. Taken together, they support the following general rules:
• If a reaction produces more gas molecules than it consumes [Example 18.2(a)],
DS 8 is positive.
• If the total number of gas molecules diminishes [Example 18.2(b)], DS8 is negative.
• If there is no net change in the total number of gas molecules [Example 18.2(c)],
then DS 8 may be positive or negative, but will be relatively small numerically.
These conclusions make sense, given that gases invariably have greater entropy than
liquids and solids. For reactions involving only liquids and solids, predicting the sign
of DS 8 is more difficult, but in many such cases an increase in the total number of
molecules and/or ions is accompanied by an increase in entropy.
18.4 The Second Law of Thermodynamics 619
We omit the subscript rxn for simplicity.
Similar problems: 18.13 and 18.14.
Predict whether the entropy change of the system in each of the following reactions is
positive or negative.
(a)
(b)
(c)
Strategy We are asked to predict, not calculate, the sign of entropy change in the
reactions. The factors that lead to an increase in entropy are (1) a transition from a
condensed phase to the vapor phase and (2) a reaction that produces more product
molecules than reactant molecules in the same phase. It is also important to compare
the relative complexity of the product and reactant molecules. In general, the more
complex the molecular structure, the greater the entropy of the compound.
Solution (a) Two reactant molecules combine to form one product molecule. Even
though H2O is a more complex molecule than either H2 and O2, the fact that there
is a net decrease of one molecule and gases are converted to liquid ensures that the
number of microstates will be diminished and hence DS 8 is negative.
(b) A solid is converted to two gaseous products. Therefore, DS 8 is positive.
(c) The same number of molecules is involved in the reactants as in the product. Fur-
thermore, all molecules are diatomic and therefore of similar complexity. As a
result, we cannot predict the sign of DS 8, but we know that the change must be
quite small in magnitude.
Practice Exercise Discuss qualitatively the sign of the entropy change expected for
each of the following processes:
(a)
(b)
(c) N2(g) 1 O2(g)¡ 2NO(g)
2Zn(s) 1 O2(g)¡ 2ZnO(s)
I2(s)¡ 2I(g)
H2(g) 1 Br2(g)¡ 2HBr(g)
NH4Cl(s)¡ NH3(g) 1 HCl(g)
2H2(g) 1 O2(g)¡ 2H2O(l )
Example 18.3
Entropy Changes in the Surroundings
Next we see how DSsurr is calculated. When an exothermic process takes place in
the system, the heat transferred to the surroundings enhances motion of the mole-
cules in the surroundings. Consequently, there is an increase in the number of
microstates and the entropy of the surroundings increases. Conversely, an endother-
mic process in the system absorbs heat from the surroundings and so decreases the
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entropy of the surroundings because molecular motion decreases (Figure 18.5). For
constant-pressure processes, the heat change is equal to the enthalpy change of the
system, DHsys. Therefore, the change in entropy of the surroundings, DSsurr, is pro-
portional to DHsys:
The minus sign is used because if the process is exothermic, DHsys is negative and
DSsurr is a positive quantity, indicating an increase in entropy. On the other hand, for
an endothermic process, DHsys is positive and the negative sign ensures that the
entropy of the surroundings decreases.
The change in entropy for a given amount of heat absorbed also depends on the
temperature. If the temperature of the surroundings is high, the molecules are already
quite energetic. Therefore, the absorption of heat from an exothermic process in the
system will have relatively little impact on molecular motion and the resulting
increase in entropy of the surroundings will be small. However, if the temperature
of the surroundings is low, then the addition of the same amount of heat will cause
a more drastic increase in molecular motion and hence a larger increase in entropy.
By analogy, someone coughing in a crowded restaurant will not disturb too many
people, but someone coughing in a library definitely will. From the inverse rela-
tionship between DSsurr and temperature T (in kelvins)—that is, the higher the tem-
perature, the smaller the DSsurr and vice versa—we can rewrite the preceding
relationship as
(18.8)
Let us now apply the procedure for calculating DSsys and DSsurr to the synthesis
of ammonia and ask whether the reaction is spontaneous at 258C:
¢H°rxn 5 292.6 kJ/molN2(g) 1 3H2(g)¡ 2NH3(g)
¢Ssurr 52¢Hsys
T
¢Ssurr r 2¢Hsys
620 CHAPTER 18 Thermodynamics
Entr
op
y d
ecre
ases
Surroundings
Heat
System
(b)
Surroundings
Heat
(a)
System
Entr
op
y i
ncr
ease
s
Figure 18.5(a) An exothermic process transfers heat from the system to the surroundings and results in an increase in the entropy of the sur-roundings. (b) An endothermic process absorbs heat from the surroundings and thereby decreases the entropy of the surroundings.
This equation, which can be derived from
the laws of thermodynamics, assumes that
both the system and the surroundings are
at temperature T.
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From Example 18.2(b) we have DSsys 5 2199 J/K ? mol, and substituting DHsys
(292.6 kJ/mol) in Equation (18.8), we obtain
The change in entropy of the universe is
Because DSuniv is positive, we predict that the reaction is spontaneous at 258C. It is
important to keep in mind that just because a reaction is spontaneous does not mean
that it will occur at an observable rate. The synthesis of ammonia is, in fact, extremely
slow at room temperature. Thermodynamics can tell us whether a reaction will occur
spontaneously under specific conditions, but it does not say how fast it will occur.
Reaction rates are the subject of chemical kinetics (see Chapter 14).
The Third Law of Thermodynamics and Absolute Entropy
Finally, it is appropriate to consider the third law of thermodynamics briefly in con-
nection with the determination of entropy values. So far we have related entropy to
microstates—the greater the number of microstates a system possesses, the larger is
the entropy of the system. Consider a perfect crystalline substance at absolute zero
(0 K). Under these conditions, molecular motions are kept at a minimum and the num-
ber of microstates (W) is one (there is only one way to arrange the atoms or mole-
cules to form a perfect crystal). From Equation (18.1) we write
According to the third law of thermodynamics, the entropy of a perfect crystalline
substance is zero at the absolute zero of temperature. As the temperature increases,
the freedom of motion increases and hence also the number of microstates. Thus, the
entropy of any substance at a temperature above 0 K is greater than zero. Note also
that if the crystal is impure or if it has defects, then its entropy is greater than zero
even at 0 K because it would not be perfectly ordered and the number of microstates
would be greater than one.
The important point about the third law of thermodynamics is that it enables us
to determine the absolute entropies of substances. Starting with the knowledge that
the entropy of a pure crystalline substance is zero at absolute zero, we can measure
the increase in entropy of the substance when it is heated from 0 K to, say, 298 K.
The change in entropy, DS, is given by
because Si is zero. The entropy of the substance at 298 K, then, is given by DS or
Sf, which is called the absolute entropy because this is the true value and not a
value derived using some arbitrary reference as in the case of standard enthalpy of
formation. Thus, the entropy values quoted so far and those listed in Appendix 2
5 Sf
¢S 5 Sf 2 Si
5 k ln 1 5 0
S 5 k ln W
5 112 J/K ? mol
5 2199 J/K ? mol 1 311 J/K ? mol
¢Suniv 5 ¢Ssys 1 ¢Ssurr
¢Ssurr 52(292.6 3 1000) J/mol
298 K5 311 J/K ? mol
18.4 The Second Law of Thermodynamics 621
34
Interactivity:Entropy vs. TemperatureARIS, Interactives
The entropy increase can be calculated
from the temperature change and heat
capacity of the substance, plus any phase
changes.
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are all absolute entropies. Because measurements are carried out at 1 atm, we usu-
ally refer to absolute entropies as standard entropies. In contrast, we cannot have
the absolute energy or enthalpy of a substance because the zero of energy or
enthalpy is undefined. Figure 18.6 shows the change (increase) in entropy of a sub-
stance with temperature. At absolute zero, it has a zero entropy value (assuming
that it is a perfect crystalline substance). As it is heated, its entropy increases grad-
ually because of greater molecular motion. At the melting point, there is a sizable
increase in entropy as the liquid state is formed. Further heating increases the
entropy of the liquid again due to enhanced molecular motion. At the boiling point
there is a large increase in entropy as a result of the liquid-to-vapor transition.
Beyond that temperature, the entropy of the gas continues to rise with increasing
temperature.
18.5 Gibbs Free Energy
The second law of thermodynamics tells us that a spontaneous reaction increases the
entropy of the universe; that is, DSuniv . 0. In order to determine the sign of DSuniv
for a reaction, however, we would need to calculate both DSsys and DSsurr. In general,
we are usually concerned only with what happens in a particular system. Therefore,
we need another thermodynamic function to help us determine whether a reaction will
occur spontaneously if we consider only the system itself.
From Equation (18.4), we know that for a spontaneous process, we have
¢Suniv 5 ¢Ssys 1 ¢Ssurr 7 0
622 CHAPTER 18 Thermodynamics
S° (
J/K
•m
ol)
Temperature (K)
Boiling(∆Svap)
Melting(∆Sfus)
Gas
Solid
Liquid
Figure 18.6Entropy increase of a substanceas the temperature rises fromabsolute zero.
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Substituting 2DHsys/T for DSsurr, we write
Multiplying both sides of the equation by T gives
Now we have a criterion for a spontaneous reaction that is expressed only in terms
of the properties of the system (DHsys and DSsys) and we can ignore the surround-
ings. For convenience, we can change the preceding equation by multiplying it
throughout by 21 and replacing the . sign with ,:
This equation says that for a process carried out at constant pressure and temperature
T, if the changes in enthalpy and entropy of the system are such that DHsys 2 TDSsys
is less than zero, the process must be spontaneous.
In order to express the spontaneity of a reaction more directly, we introduce
another thermodynamic function called Gibbs free energy (G), or simply free energy
(after the American physicist Josiah Willard Gibbs):
(18.9)
All quantities in Equation (18.9) pertain to the system, and T is the temperature of
the system. You can see that G has units of energy (both H and TS are in energy
units). Like H and S, G is a state function.
The change in free energy (DG) of a system for a constant-temperature process is
(18.10)
In this context free energy is the energy available to do work. Thus, if a particular
reaction is accompanied by a release of usable energy (that is, if DG is negative), this
fact alone guarantees that it is spontaneous, and there is no need to worry about what
happens to the rest of the universe.
Note that we have merely organized the expression for the entropy change of the
universe and equating the free-energy change of the system (DG) with 2TDSuniv, so
that we can focus on changes in the system. We can now summarize the conditions
for spontaneity and equilibrium at constant temperature and pressure in terms of DG
as follows:
The reaction is spontaneous in the forward direction.
The reaction is nonspontaneous. The reaction is
spontaneous in the opposite direction.
The system is at equilibrium. There is no net change.
Standard Free-Energy Changes
The standard free-energy of reaction (DG8rxn) is the free-energy change for a reac-
tion when it occurs under standard-state conditions, when reactants in their standard
states are converted to products in their standard states. Table 18.2 summarizes the
¢G 5 0
¢G 7 0
¢G 6 0
¢G 5 ¢H 2 T¢S
G 5 H 2 TS
2T¢Suniv 5 ¢Hsys 2 T¢Ssys 6 0
T¢Suniv 5 2¢Hsys 1 T¢Ssys 7 0
¢Suniv 5 ¢Ssys 2¢Hsys
T7 0
18.5 Gibbs Free Energy 623
The change in unequal sign when we
multiply the equation by 1 follows from
the fact that 1 . 0 and 1 , 0.
The word “free” in the term “free
energy” does not mean without cost.
A commemorative stamp honor-ing Gibbs.
We omit the subscript sys for simplicity.
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conventions used by chemists to define the standard states of pure substances as well
as solutions. To calculate (DG 8rxn) we start with the equation
The standard free-energy change for this reaction is given by
(18.11)
or, in general,
(18.12)
where m and n are stoichiometric coefficients. The term DG8f is the standard free
energy of formation of a compound, that is, the free-energy change that occurs when
1 mole of the compound is synthesized from its elements in their standard states. For
the combustion of graphite:
the standard free-energy change [from Equation (18.12)] is
As in the case of the standard enthalpy of formation (p. 192), we define the standard
free energy of formation of any element in its stable allotropic form at 1 atm and 258C
as zero. Thus,
and
Therefore, the standard free-energy change for the reaction in this case is equal to the
standard free energy of formation of CO2:
Appendix 2 lists the values of DG 8f for a number of compounds.
¢G°rxn 5 ¢G°f (CO2)
¢G°f (O2) 5 0¢G°f (C, graphite) 5 0
¢G°rxn 5 ¢G°f (CO2) 2 [¢G°f (C, graphite) 1 ¢G°f (O2)]
C(graphite) 1 O2(g)¡ CO2(g)
¢G°rxn 5 ©n¢G°f (products) 2 ©m¢G°f (reactants)
¢G°rxn 5 [c¢G°f (C) 1 d¢G°f (D)] 2 [a¢G°f (A) 1 b¢G°f (B)]
aA 1 bB¡ cC 1 dD
624 CHAPTER 18 Thermodynamics
State of Standard Matter State
Gas 1 atm pressure
Liquid Pure liquid
Solid Pure solid
Elements* DG8f 5 0
Solution 1 molar
concentration
TABLE 18.2
Conventions for
Standard States
*The most stable allotropic form at 258C
and 1 atm.
Calculate the standard free-energy changes for the following reactions at 258C.
(a)
(b)
Strategy To calculate the standard free-energy change of a reaction, we look up the
standard free energies of formation of reactants and products in Appendix 2 and apply
Equation (18.12). Note that all the stoichiometric coefficients have no units so DG8rxn is
expressed in units of kJ/mol, and DG8f for O2 is zero because it is the stable allotropic
element at 1 atm and 258C.
Solution (a) According to Equation (18.12), we write
(Continued )
¢G°rxn 5 [¢G°f (CO2) 1 2¢G°f (H2O)] 2 [¢G°f (CH4) 1 2¢G°f (O2)]
2MgO(s)¡ 2Mg(s) 1 O2(g)
CH4(g) 1 2O2(g)¡ CO2(g) 1 2H2O(l)
Example 18.4
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Applications of Equation (18.10)
In order to predict the sign of DG, according to Equation (18.10) we need to know
both DH and DS. A negative DH (an exothermic reaction) and a positive DS (a reac-
tion that results in an increase in the microstates of the system) tend to make DG neg-
ative, although temperature may also influence the direction of a spontaneous reaction.
The four possible outcomes of this relationship are:
• If both DH and DS are positive, then DG will be negative only when the TDS
term is greater in magnitude than DH. This condition is met when T is large.
• If DH is positive and DS is negative, DG will always be positive, regardless of
temperature.
• If DH is negative and DS is positive, then DG will always be negative regardless
of temperature.
• If DH is negative and DS is negative, then DG will be negative only when TDS
is smaller in magnitude than DH. This condition is met when T is small.
The temperatures that will cause DG to be negative for the first and last cases depend
on the actual values of DH and DS of the system. Table 18.3 summarizes the effects
of the possibilities just described.
Before we apply the change in free energy to predict reaction spontaneity, it is
useful to distinguish between DG and DG 8. Suppose we carry out a reaction in solu-
tion with all the reactants in their standard states (that is, all at 1 M concentration).
As soon as the reaction starts, the standard-state condition no longer exists for the
reactants or the products because their concentrations are different from 1 M. Under
nonstandard state conditions, we must use the sign of DG rather than that of DG8 to
predict the direction of the reaction. The sign of DG 8, on the other hand, tells us
whether the products or the reactants are favored when the reacting system reaches
equilibrium. Thus, a negative value of DG8 indicates that the reaction favors product
formation whereas a positive value of DG 8 indicates that there will be more reactants
than products at equilibrium.
18.5 Gibbs Free Energy 625
Similar problems: 18.17 and 18.18.
We insert the appropriate values from Appendix 2:
(b) The equation is
From data in Appendix 2 we write
Practice Exercise Calculate the standard free-energy changes for the following
reactions at 258C:
(a)
(b) 2C2H6(g) 1 7O2(g)¡ 4CO2(g) 1 6H2O(l)
H2(g) 1 Br2(l)¡ 2HBr(g)
5 1139 kJ/mol
¢G°rxn 5 [(2)(0 kJ/mol) 1 (0 kJ/mol)] 2 [(2)(2569.6 kJ/mol)]
¢G°rxn 5 [2¢G°f (Mg) 1 ¢G°f (O2)] 2 [2¢G°f (MgO)]
5 2818.0 kJ/mol
[(250.8 kJ/mol) 1 (2)(0 kJ/mol)]
¢G°rxn 5 [(2394.4 kJ/mol) 1 (2)(2237.2 kJ/mol)] 2
In Section 18.6 we will see an equation
relating DG8 to the equilibrium constant K.
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We will now consider two specific applications of Equation (18.10).
Temperature and Chemical Reactions
Calcium oxide (CaO), also called quicklime, is an extremely valuable inorganic sub-
stance used in steelmaking, production of calcium metal, the paper industry, water
treatment, and pollution control. It is prepared by decomposing limestone (CaCO3) in
a kiln at a high temperature (Figure 18.7):
The reaction is reversible, and CaO readily combines with CO2 to form CaCO3. The
pressure of CO2 in equilibrium with CaCO3 and CaO increases with temperature. In
the industrial preparation of quicklime, the system is never maintained at equilibrium;
rather, CO2 is constantly removed from the kiln to shift the equilibrium from left to
right, promoting the formation of calcium oxide.
The important information for the practical chemist is the temperature at which
the decomposition of CaCO3 becomes appreciable (that is, the temperature at which
the reaction begins to favor products). We can make a reliable estimate of that tem-
perature as follows. First we calculate DH8 and DS 8 for the reaction at 258C, using
the data in Appendix 2. To determine DH 8 we apply Equation (6.17):
Next we apply Equation (18.6) to find DS8
From Equation (18.10)
we obtain
5 130.0 kJ/mol
¢G° 5 177.8 kJ/mol 2 (298 K)(160.5 J/K ? mol)a 1 kJ
1000 Jb
¢G° 5 ¢H° 2 T¢S°
5 160.5 J/K ? mol
5 [(39.8 J/K ? mol) 1 (213.6 J/K ? mol)] 2 (92.9 J/K ? mol)
¢S° 5 [S°(CaO) 1 S°(CO2)] 2 S°(CaCO3)
5 177.8 kJ/mol
5 [(2635.6 kJ/mol) 1 (2393.5 kJ/mol)] 2 (21206.9 kJ/mol)
¢H° 5 [¢H°f (CaO) 1 ¢H°f (CO2)] 2 [¢H°f (CaCO3)]
CaCO3(s)∆ CaO(s) 1 CO2(g)
626 CHAPTER 18 Thermodynamics
DH DS Example
1 1 Reaction proceeds spontaneously at high temperatures. At low
temperatures, reaction is spontaneous in the reverse direction.
1 2 DG is always positive. Reaction is spontaneous in the reverse
direction at all temperatures.
2 1 DG is always negative. Reaction proceeds spontaneously at
all temperatures.
2 2 Reaction proceeds spontaneously at low temperatures. At high
temperatures, the reverse reaction becomes spontaneous.
NH3(g) 1 HCl(g)¡ NH4Cl(s)
2H2O2(l)¡ 2H2O(l) 1 O2(g)
3O2(g)¡ 2O3(g)
2HgO(s)¡ 2Hg(l) 1 O2(g)
DG
TABLE 18.3
Le Châtelier’s principle predicts that the
forward, endothermic reaction is favored
by heating.
Figure 18.7The production of CaO fromCaCO3 in a rotatory kiln.
Factors Affecting the Sign of DG in the Relationship DG 5 DH TDS
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Because DG8 is a large positive quantity, we conclude that the reaction is not favored
for product formation at 258C (or 298 K). Indeed, the pressure of CO2 is so low at
room temperature that it cannot be measured. In order to make DG 8 negative, we first
have to find the temperature at which DG8 is zero; that is,
or
At a temperature higher than 8358C, DG8 becomes negative, indicating that the
reaction now favors the formation of CaO and CO2. For example, at 8408C, or
1113 K,
Two points are worth making about such a calculation. First, we used the DH8
and DS8 values at 258C to calculate changes that occur at a much higher temperature.
Because both DH8 and DS8 change with temperature, this approach will not give us
an accurate value of DG8, but it is good enough for “ball park” estimates. Second, we
should not be misled into thinking that nothing happens below 8358C and that at 8358C
CaCO3 suddenly begins to decompose. Far from it. The fact that DG8 is a positive
value at some temperature below 8358C does not mean that no CO2 is produced, but
rather that the pressure of the CO2 gas formed at that temperature will be below 1 atm
(its standard-state value; see Table 18.2). As Figure 18.8 shows, the pressure of CO2
at first increases very slowly with temperature; it becomes easily measurable above
7008C. The significance of 8358C is that this is the temperature at which the equilib-
rium pressure of CO2 reaches 1 atm. Above 8358C, the equilibrium pressure of CO2
exceeds 1 atm.
Phase Transitions
At the temperature at which a phase transition occurs (that is, at the melting point or
boiling point) the system is at equilibrium (DG 5 0), so Equation (18.10) becomes
or
Let us first consider the ice-water equilibrium. For the ice n water transition, DH is
the molar heat of fusion (see Table 12.7), and T is the melting point. The entropy
change is therefore
5 22.0 J/K ? mol
¢SiceSwater 56010 J/mol
273 K
¢S 5¢H
T
0 5 ¢H 2 T¢S
¢G 5 ¢H 2 T¢S
5 20.8 kJ/mol
5 177.8 kJ/mol 2 (1113 K)(160.5 J/K ? mol)a 1 kJ
1000 Jb
¢G° 5 ¢H° 2 T¢S°
5 1108 K or 835°C
5(177.8 kJ/mol)(1000 J/1 kJ)
160.5 J/K ? mol
T 5¢H°
¢S°
0 5 ¢H° 2 T¢S°
18.5 Gibbs Free Energy 627
The equilibrium constant of this reaction
is .KP 5 PCO2
Figure 18.8Equilibrium pressure of CO2
from the decomposition ofCaCO3, as a function of tem-perature. This curve is calcu-lated by assuming that DH8 andDS 8 of the reaction do notchange with temperature.
PC
O2(a
tm)
200
t (°C)
0
1
2
3
835°C
400 600 800
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Thus, when 1 mole of ice melts at 08C, there is an increase in entropy of 22.0 J/K ? mol.
The increase in entropy is consistent with the increase in microstates from solid to
liquid. Conversely, for the water n ice transition, the decrease in entropy is
given by
In the laboratory we normally carry out unidirectional changes, that is, either ice to
water or water to ice transition. We can calculate entropy change in each case using
the equation DS 5 DHyT as long as the temperature remains at 08C. The same pro-
cedure can be applied to the water n steam transition. In this case DH is the heat of
vaporization and T is the boiling point of water.
5 222.0 J/K ? mol
¢SwaterSice 526010 J/mol
273 K
628 CHAPTER 18 Thermodynamics
The melting of ice is an endothermic
process (DH is positive), and the freezing
of water is exothermic (DH is negative).
Liquid and solid benzene inequilibrium at 5.58C.
Similar problem: 18.60.
The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol,
respectively. Calculate the entropy changes for the solid n liquid and liquid n
vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.58C and boils
at 80.18C.
Strategy At the melting point, liquid and solid benzene are at equilibrium, so
DG 5 0. From Equation (18.10) we have DG 5 0 5 DH 2 TDS or DS 5 DH/T. To
calculate the entropy change for the solid benzene n liquid benzene transition, we
write DSfus 5 DHfusyTf. Here DHfus is positive for an endothermic process, so DSfus
is also positive, as expected for a solid to liquid transition. The same procedure
applies to the liquid benzene n vapor benzene transition. What temperature unit
should be used?
Solution The entropy change for melting 1 mole of benzene at 5.58C is
Similarly, the entropy change for boiling 1 mole of benzene at 80.18C is
Check Because vaporization creates more microstates than the melting process,
DSvap . DSfus.
Practice Exercise The molar heats of fusion and vaporization of argon are 1.3 kJ/mol
and 6.3 kJ/mol, and argon’s melting point and boiling point are 21908C and 21868C,
respectively. Calculate the entropy changes for fusion and vaporization.
5 87.8 J/K ? mol
5(31.0 kJ/mol)(1000 J/1 kJ)
(80.1 1 273) K
¢Svap 5¢Hvap
Tbp
5 39.1 J/K ? mol
5(10.9 kJ/mol)(1000 J/1 kJ)
(5.5 1 273) K
¢Sfus 5¢Hfus
¢Tf
Example 18.5
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18.6 Free Energy and Chemical Equilibrium
As mentioned earlier, during the course of a chemical reaction not all the reactants
and products will be at their standard states. Under this condition, the relationship
between DG and DG8, which can be derived from thermodynamics, is
(18.13)
where R is the gas constant (8.314 J/K ? mol), T is the absolute temperature of the
reaction, and Q is the reaction quotient (see p. 507). We see that DG depends on two
quantities: DG8 and RT ln Q. For a given reaction at temperature T the value of DG8
is fixed but that of RT ln Q is not, because Q varies according to the composition of
the reaction mixture. Let us consider two special cases:
Case 1: A large negative value of DG8 will tend to make DG also negative. Thus, the
net reaction will proceed from left to right until a significant amount of product
has been formed. At that point, the RT ln Q term will become positive enough to
match the negative DG8 term.
Case 2: A large positive DG8 term will tend to make DG also positive. Thus, the net
reaction will proceed from right to left until a significant amount of reactant has
been formed. At that point, the RT ln Q term will become negative enough to
match the positive DG8 term.
At equilibrium, by definition, DG 5 0 and Q 5 K, where K is the equilibrium
constant. Thus,
or
(18.14)
In this equation, KP is used for gases and Kc for reactions in solution. Note that the
larger the K is, the more negative DG8 is. For chemists, Equation (18.14) is one of
the most important equations in thermodynamics because it enables us to find the
equilibrium constant of a reaction if we know the change in standard free energy and
vice versa.
It is significant that Equation (18.14) relates the equilibrium constant to the
standard free energy change DG8 rather than to the actual free energy change DG.
The actual free energy change of the system varies as the reaction progresses and
becomes zero at equilibrium. On the other hand, DG8 is a constant for a particular
reaction at a given temperature. Figure 18.9 shows plots of the free energy of a react-
ing system versus the extent of the reaction for two types of reactions. As you can
see, if DG8 , 0, the products are favored over reactants at equilibrium. Conversely,
if DG8 . 0, there will be more reactants than products at equilibrium. Table 18.4 sum-
marizes the three possible relations between DG8 and K, as predicted by Equation
(18.14). Remember this important distinction: It is the sign of DG and not that of DG8
that determines the direction of reaction spontaneity. The sign of DG8 tells us only
the relative amounts of products and reactants when equilibrium is reached, not the
direction of the net reaction.
For reactions having very large or very small equilibrium constants, it is gener-
ally very difficult, if not impossible, to measure the K values by monitoring the
¢G° 5 2RT ln K
0 5 ¢G° 1 RT ln K
¢G 5 ¢G° 1 RT ln Q
18.6 Free Energy and Chemical Equilibrium 629
Sooner or later a reversible reaction will
reach equilibrium.
Interactivity:Free Energy—EquilibriumARIS, Interactives
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concentrations of all the reacting species. Consider, for example, the formation of
nitric oxide from molecular nitrogen and molecular oxygen:
At 258C, the equilibrium constant KP is
The very small value of KP means that the concentration of NO at equilibrium will
be exceedingly low. In such a case the equilibrium constant is more conveniently
obtained from DG8. (As we have seen, DG 8 can be calculated from DH8 and DS 8.)
On the other hand, the equilibrium constant for the formation of hydrogen iodide from
molecular hydrogen and molecular iodine is near unity at room temperature:
For this reaction it is easier to measure KP and then calculate DG8 using Equation
(18.14) than to measure DH 8 and DS 8 and use Equation (18.10).
H2(g) 1 I2(g)∆ 2HI(g)
KP 5P2
NO
PN2PO2
5 4.0 3 10231
N2(g) 1 O2(g)∆ 2NO(g)
630 CHAPTER 18 Thermodynamics
Figure 18.9(a) DG 8 , 0. At equilibrium, there is a significant conversion of reactants to products. (b) DG 8 . 0. At equilibrium, reactantsare favored over products. In both cases, the net reaction toward equilibrium is from left to right (reactants to products) if Q , K and right to left (products to reactants) if Q . K.
Extent of reaction
Purereactants
Equilibriumposition
Equilibriumposition
Pureproducts
G° 5 G°(products) –
Q 5 K Q 5 K
Q < K Q > K
Q > K Q < K
∆
G°(reactants) < 0
Fre
e en
ergy (
G)
of
the
reac
ting s
yst
em
G°(reactants)
G°(products)
(a)
Extent of reaction
Purereactants
Pureproducts
Fre
e en
ergy (
G)
of
the
reac
ting s
yst
em
G°(reactants)
G°(products)
(b)
G° 5 G°(products) –∆
G°(reactants) > 0
K ln K Comments
. 1 Positive Negative Products are favored over reactants at equilibrium.
5 1 0 0 Products and reactants are equally favored at
equilibrium.
, 1 Negative Positive Reactants are favored over products at equilibrium.
DG8
Relation Between DG 8 and K as Predicted by the
Equation DG 8 5 RT ln K
TABLE 18.4
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18.6 Free Energy and Chemical Equilibrium 631
Similar problems: 18.23 and 18.26.
To calculate KP, enter 2191.5 on your
calculator and then press the key labeled
“e” or “inv(erse) ln x.”
Using data listed in Appendix 2, calculate the equilibrium constant (KP) for the follow-
ing reaction at 258C:
Strategy According to Equation (18.14), the equilibrium constant for the reaction is
related to the standard free energy change; that is, DG 8 5 2RT ln K. Therefore, we
first need to calculate DG 8 by following the procedure in Example 18.4. Then we can
calculate KP. What temperature unit should be used?
Solution According to Equation (18.12),
Using Equation (18.14)
Comment This extremely small equilibrium constant is consistent with the fact that
water does not decompose into hydrogen and oxygen gases at 258C. Thus, a large
positive DG 8 favors reactants over products at equilibrium.
Practice Exercise Calculate the equilibrium constant (KP) for the reaction at 258C
2O3(g)¡ 3O2(g)
KP 5 e2191.55 7 3 10284
ln KP 5 2191.5
474.4 kJ/mol 31000 J
1 kJ5 2(8.314 J/K # mol)(298 K) ln KP
¢G°rxn 5 2RT ln KP
5 474.4 kJ/mol
5 [(2)(0 kJ/mol) 1 (0 kJ/mol)] 2 [(2)(2237.2 kJ/mol)]
¢G°rxn 5 [2¢G°f (H2) 1 ¢G°f (O2)] 2 [2¢G°f (H2O)]
2H2O(l)∆ 2H2(g) 1 O2(g)
In Chapter 17 we discussed the solubility product of slightly soluble substances. Using the
solubility product of silver chloride at 258C (1.6 3 10210), calculate DG8 for the process
Strategy According to Equation (18.14), the equilibrium constant for the reaction is
related to standard free energy change; that is, DG 8 5 2RT ln K. Because this is a
heterogeneous equilibrium, the solubility product (Ksp) is the equilibrium constant. We
calculate the standard free energy change from the Ksp value of AgCl. What tempera-
ture unit should be used?
Solution The solubility equilibrium for AgCl is
Using Equation (18.14) we obtain
(Continued )
5 56 kJ/mol
5 5.6 3 104 J/mol
¢G° 5 2(8.314 J/K # mol)(298 K) ln (1.6 3 10210)
Ksp 5 [Ag1][Cl2] 5 1.6 3 10210
AgCl(s)∆ Ag1(aq) 1 Cl2(aq)
AgCl(s)∆ Ag1(aq) 1 Cl2(aq)
Example 18.6
Example 18.7
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18.7 Thermodynamics in Living Systems
Many biochemical reactions have a positive DG 8 value, yet they are essential to the
maintenance of life. In living systems these reactions are coupled to an energetically
favorable process, one that has a negative DG 8 value. The principle of coupled
632 CHAPTER 18 Thermodynamics
Similar problem: 18.25.
Check The large, positive DG 8 indicates that AgCl is slightly soluble and that the
equilibrium lies mostly to the left.
Practice Exercise Calculate DG 8 for the following process at 258C:
The Ksp of BaF2 is 1.7 3 1026.
BaF2(s)∆ Ba21(aq) 1 2F2(aq)
Similar problems: 18.27 and 18.28.
The equilibrium constant (KP) for the reaction
is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.40 kJ/mol. In
a certain experiment, the initial pressures are PNO25 0.122 atm and PN2O4
5 0.453 atm.
Calculate DG for the reaction at these pressures and predict the direction of the net
reaction.
Strategy From the information given we see that neither the reactant nor the product
is at its standard state of 1 atm. To determine the direction of the net reaction, we need
to calculate the free-energy change under nonstandard-state conditions (DG) using
Equation (18.13) and the given DG 8 value. Note that the partial pressures are expressed
as dimensionless quantities in the reaction quotient QP.
Solution Equation (18.13) can be written as
Because DG , 0, the net reaction proceeds from left to right to reach equilibrium.
Check Note that although DG 8 . 0, the reaction can be made to favor product
formation initially by having a small concentration (pressure) of the product compared
to that of the reactant. Confirm the prediction by showing that QP , KP.
Practice Exercise The DG 8 for the reaction
is 2.60 kJ/mol at 258C. In one experiment, the initial pressures are PH25 4.26 atm,
, and PHI 5 0.23 atm. Calculate DG for the reaction and predict the
direction of the net reaction.
PI25 0.024 atm
H2(g) 1 I2(g)∆ 2HI(g)
5 23.06 3 103 J/mol 5 23.06 kJ/mol
5 5.40 3 103 J/mol 2 8.46 3 103 J/mol
5 5.40 3 103 J/mol 1 (8.314 J/K ? mol)(298 K) 3 ln(0.122)2
0.453
5 ¢G° 1 RT ln PNO2
2
PN2O4
¢G 5 ¢G° 1 RT ln QP
N2O4(g)∆ 2NO2(g)
Example 18.8
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reactions is based on a simple concept: we can use a thermodynamically favorable
reaction to drive an unfavorable one. Consider an industrial process. Suppose we wish
to extract zinc from the ore sphalerite (ZnS). The following reaction will not work
because it has a large positive DG8 value:
On the other hand, the combustion of sulfur to form sulfur dioxide is favored because
of its large negative DG8 value:
By coupling the two processes we can bring about the separation of zinc from zinc
sulfide. In practice, this means heating ZnS in air so that the tendency of S to form
SO2 will promote the decomposition of ZnS:
Coupled reactions play a crucial role in our survival. In biological systems,
enzymes facilitate a wide variety of nonspontaneous reactions. For example, in the
human body, food molecules, represented by glucose (C6H12O6), are converted to car-
bon dioxide and water during metabolism with a substantial release of free energy:
In a living cell, this reaction does not take place in a single step (as burning glucose
in a flame would); rather, the glucose molecule is broken down with the aid of
enzymes in a series of steps. Much of the free energy released along the way is used
to synthesize adenosine triphosphate (ATP) from adenosine diphosphate (ADP) and
phosphoric acid (Figure 18.10):
The function of ATP is to store free energy until it is needed by cells. Under appro-
priate conditions, ATP undergoes hydrolysis to give ADP and phosphoric acid, with
a release of 31 kJ of free energy, which can be used to drive energetically unfavor-
able reactions, such as protein synthesis.
Proteins are polymers made of amino acids. The stepwise synthesis of a protein
molecule involves the joining of individual amino acids. Consider the formation of
¢G° 5 131 kJ/molADP 1 H3PO4 ¡ ATP 1 H2O
¢G° 5 22880 kJ/molC6H12O6(s) 1 6O2(g)¡ 6CO2(g) 1 6H2O(l)
¢G° 5 2101.8 kJ/molZnS(s) 1 O2(g)¡ Zn(s) 1 SO2(g)
¢G° 5 2300.1 kJ/molS(s) 1 O2(g)¡ SO2(g)
¢G° 5 198.3 kJ/molZnS(s)¡ Zn(s) 1 S(s)
¢G° 5 2300.1 kJ/molS(s) 1 O2(g)¡ SO2(g)
¢G° 5 198.3 kJ/molZnS(s)¡ Zn(s) 1 S(s)
18.7 Thermodynamics in Living Systems 633
A mechanical analog for coupledreactions. We can make thesmaller weight move upward (anonspontaneous process) bycoupling it with the falling of alarger weight.
O B
A 2O
O B
A 2O
O B
A 2O
2OOPOOOPOOOPOOCH2
NC
HCCH
NH2
C
CN
N
N
Adenosine triphosphate(ATP)
OAAH
HAA
HO
HAAOH
AAAAAH
O B
A 2O
O B
A 2O
2OOPOOOPOOCH2
NC
HCCH
NH2
C
CN
N
N
Adenosine diphosphate(ADP)
OAAH
HAA
HO
HAAOH
AAAAAH
Figure 18.10Structure of ATP and ADP inionized forms. The adeninegroup is in blue, the ribosegroup in black, and the phos-phate group in red. Note thatADP has one fewer phosphategroup than ATP.
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the dipeptide (a two-amino-acid unit) alanylglycine from alanine and glycine. This
reaction represents the first step in the synthesis of a protein molecule:
As you can see, this reaction does not favor the formation of product, and so only a
little of the dipeptide would be formed at equilibrium. However, with the aid of an
enzyme, the reaction is coupled to the hydrolysis of ATP as follows:
The overall free-energy change is given by DG8 5 231 kJ/mol 1 29 kJ/mol 5 22 kJ/mol,
which means that the coupled reaction now favors the formation of product, and an
appreciable amount of alanylglycine will be formed under this condition. Figure 18.11
shows the ATP-ADP interconversions that act as energy storage (from metabolism) and
free energy release (from ATP hydrolysis) to drive essential reactions.
ATP 1 H2O 1 Alanine 1 Glycine¡ ADP 1 H3PO4 1 Alanylglycine
¢G° 5 129 kJ/molAlanine 1 Glycine¡ Alanylglycine
634 CHAPTER 18 Thermodynamics
ATPGlucose
Amino acids
Proteins
ADP
Fre
e E
ner
gy
CO2 1 H2O
Figure 18.11Schematic representation ofATP synthesis and coupledreactions in living systems. Theconversion of glucose to carbondioxide and water duringmetabolism releases free energy.The released free energy is usedto convert ADP into ATP. TheATP molecules are then used asan energy source to drive unfa-vorable reactions, such asprotein synthesis from aminoacids.
KEY EQUATIONS
S 5 k ln W (18.1) Relating entropy to number of microstates.
DSuniv 5 DSsys 1 DSsurr . 0 (18.4) The second law of thermodynamics (spontaneous
process).
DSuniv 5 DSsys 1 DSsurr 5 0 (18.5) The second law of thermodynamics (equilibrium
process).
DS 8rxn 5 SnS 8(products)
2 SmS 8(reactants) (18.7) Standard entropy change of a reaction.
G 5 H 2 TS (18.9) Definition of Gibbs free energy.
DG 5 DH 2 TDS (18.10) Free-energy change at constant temperature.
DG8rxn 5 SnDG8f (products)
2 SmDG8f (reactants) (18.12) Standard free-energy change of a reaction.
DG 5 DG8 1 RT ln Q (18.13) Relationship between free-energy change and
standard free-energy change and reaction quotient.
DG8 5 2RT ln K (18.14) Relationship between standard free-energy
change and the equilibrium constant.
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Questions and Problems 635
1. Entropy is usually described as a measure of the disor-
der of a system. Any spontaneous process must lead to
a net increase in entropy in the universe (second law of
thermodynamics).
2. The standard entropy of a chemical reaction can be calcu-
lated from the absolute entropies of reactants and products.
3. The third law of thermodynamics states that the entropy of
a perfect crystalline substance is zero at 0 K. This law en-
ables us to measure the absolute entropies of substances.
4. Under conditions of constant temperature and pressure,
the free-energy change DG is less than zero for a spon-
taneous process and greater than zero for a nonsponta-
neous process. For an equilibrium process, DG 5 0.
5. For a chemical or physical process at constant tempera-
ture and pressure, DG 5 DH 2 TDS. This equation can
be used to predict the spontaneity of a process.
6. The standard free-energy change for a reaction, DG8,
can be calculated from the standard free energies of for-
mation of reactants and products.
7. The equilibrium constant of a reaction and the standard
free-energy change of the reaction are related by the
equation DG8 5 2RT ln K.
8. Many biological reactions are nonspontaneous. They
are driven by the hydrolysis of ATP, for which DG8 is
negative.
SUMMARY OF FACTS AND CONCEPTS
Entropy (S), p. 612
Free energy (G), p. 623
Gibbs free energy (G), p. 623
Third law of thermodynamics,p. 621
Second law ofthermodynamics, p. 617
Standard entropy of reaction(DS8rxn), p. 618
Standard free energy offormation (DG8f ), p. 624
Standard free energy ofreaction (DG8rxn), p. 623
KEY WORDS
QUESTIONS AND PROBLEMS
Spontaneous Processes and Entropy
Review Questions
18.1 Explain what is meant by a spontaneous process.
Give two examples each of spontaneous and non-
spontaneous processes.
18.2 Which of the following processes are spontaneous
and which are nonspontaneous? (a) dissolving
table salt (NaCl) in hot soup; (b) climbing Mt.
Everest; (c) spreading fragrance in a room by re-
moving the cap from a perfume bottle; (d) sepa-
rating helium and neon from a mixture of the
gases
18.3 Which of the following processes are spontaneous and
which are nonspontaneous at a given temperature?
(a) saturated soln
(b) unsaturated soln
(c) supersaturated soln
18.4 Define entropy. What are the units of entropy?
NaNO3(s)¡H2O NaNO3(aq)
NaNO3(s)¡H2O NaNO3(aq)
NaNO3(s)¡H2O NaNO3(aq)
18.5 How does the entropy of a system change for each of
the following processes?
(a) A solid melts.
(b) A liquid freezes.
(c) A liquid boils.
(d) A vapor is converted to a solid.
(e) A vapor condenses to a liquid.
(f) A solid sublimes.
(g) Urea dissolves in water.
Problems
18.6 Consider the situation shown in Figure 18.1(a). After
the valve is opened, the probability of finding one
molecule in either bulb is (because the bulbs have
the same volume) and that of finding the molecule in
the total volume is one. The probability of finding
two molecules in the same bulb is the product of the
individual probabilities; that is, or . Extend14
12 3
12
12
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636 CHAPTER 18 Thermodynamics
the calculation to finding 100 molecules in the same
bulb. Based on your results, explain why it is highly
improbable for the process shown in Figure 18.1(b)
to occur spontaneously when the number of mole-
cules becomes very large, say, 6 3 1023.
The Second Law of Thermodynamics
Review Questions
18.7 State the second law of thermodynamics in words
and express it mathematically.
18.8 State the third law of thermodynamics and explain
its usefulness in calculating entropy values.
Problems
18.9 For each pair of substances listed here, choose the one
having the larger standard entropy value at 258C.
The same molar amount is used in the comparison.
Explain the basis for your choice. (a) Li(s) or Li(l);
(b) C2H5OH(l) or CH3OCH3(l) (Hint: Which mol-
ecule can hydrogen-bond?); (c) Ar(g) or Xe(g);
(d) CO(g) or CO2(g); (e) O2(g) or O3(g);
(f) NO2(g) or N2O4(g)
18.10 Arrange the following substances (1 mole each) in
order of increasing entropy at 258C: (a) Ne(g),
(b) SO2(g), (c) Na(s), (d) NaCl(s), (e) H2(g). Give
the reasons for your arrangement.
18.11 Using the data in Appendix 2, calculate the standard
entropy changes for the following reactions at 258C:
(a) S(s) 1 O2(g) SO2(g)
(b) MgCO3(s) MgO(s) 1 CO2(g)
18.12 Using the data in Appendix 2, calculate the standard
entropy changes for the following reactions at 258C:
(a) H2(g) 1 CuO(s) Cu(s) 1 H2O(g)
(b) 2Al(s) 1 3ZnO(s) Al2O3(s) 1 3Zn(s)
(c) CH4(g) 1 2O2(g) CO2(g) 1 2H2O(l)
18.13 Without consulting Appendix 2, predict whether the
entropy change is positive or negative for each of the
following reactions. Give reasons for your predictions.
(a) 2KClO4(s) 2KClO3(s) 1 O2(g)
(b) H2O(g) H2O(l)
(c) 2Na(s) 1 2H2O(l ) 2NaOH(aq) 1 H2(g)
(d) N2(g) 2N(g)
18.14 State whether the sign of the entropy change ex-
pected for each of the following processes will be
positive or negative, and explain your predictions.
(a) PCl3(l) 1 Cl2(g) PCl5(s)
(b) 2HgO(s) 2Hg(l) 1 O2(g)
(c) H2(g) 2H(g)
(d) U(s) 1 3F2(g) UF6(s)¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
Gibbs Free Energy
Review Questions
18.15 Define free energy. What are its units?
18.16 Why is it more convenient to predict the direction of
a reaction in terms of DGsys instead of DSuniv? Under
what conditions can DGsys be used to predict the
spontaneity of a reaction?
Problems
18.17 Calculate DG 8 for the following reactions at 258C:
(a) N2(g) 1 O2(g) 2NO(g)
(b) H2O(l) H2O(g)
(c) 2C2H2(g) 1 5O2(g) 4CO2(g) 1 2H2O(l)
(Hint: Look up the standard free energies of forma-
tion of the reactants and products in Appendix 2.)
18.18 Calculate DG 8 for the following reactions at 258C:
(a) 2Mg(s) 1 O2(g) 2MgO(s)
(b) 2SO2(g) 1 O2(g) 2SO3(g)
(c) 2C2H6(g) 1 7O2(g) 4CO2(g) 1 6H2O(l )
See Appendix 2 for thermodynamic data.
18.19 From the values of DH and DS, predict which of the
following reactions would be spontaneous at 258C:
Reaction A: DH 5 10.5 kJ/mol, DS 5 30 J/K ? mol;
reaction B: DH 5 1.8 kJ/mol, DS 5 2113 J/K ? mol.
If either of the reactions is nonspontaneous at 258C,
at what temperature might it become spontaneous?
18.20 Find the temperatures at which reactions with the
following DH and DS values would become sponta-
neous: (a) DH 5 2126 kJ/mol, DS 5 84 J/K ? mol;
(b) DH 5 211.7 kJ/mol, DS 5 2105 J/K ? mol.
Free Energy and Chemical Equilibrium
Review Questions
18.21 Explain the difference between DG and DG 8.
18.22 Explain why Equation (18.14) is of great importance
in chemistry.
Problems
18.23 Calculate KP for the following reaction at 258C:
H2(g) 1 I2(g) 2HI(g) DG° 5 2.60 kJ/mol
18.24 For the autoionization of water at 258C,
H2O(l ) H1(aq) 1 OH2(aq)
Kw is 1.0 3 10214. What is DG° for the process?
18.25 Consider the following reaction at 258C:
Calculate DG 8 for the reaction. Ksp for Fe(OH)2 is
1.6 3 10214.
Fe(OH)2(s) ∆ Fe21(aq) 1 2OH2(aq)
∆
∆
¡
¡
¡
¡
¡
¡
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Questions and Problems 637
18.26 Calculate DG 8 and KP for the following equilibrium
reaction at 258C.
18.27 (a) Calculate DG8 and KP for the following equilib-
rium reaction at 258C. The DG°f values are 0 for
Cl2(g), 2286 kJ/mol for PCl3(g), and 2325 kJ/mol
for PCl5(g).
(b) Calculate DG for the reaction if the partial pres-
sures of the initial mixture are PPCl55 0.0029 atm,
P 5 0.27 atm, and P 5 0.40 atm.
18.28 The equilibrium constant (KP) for the reaction
is 4.40 at 2000 K. (a) Calculate DG 8 for the reaction.
(b) Calculate DG for the reaction when the partial
pressures are ,
, and .
18.29 Consider the decomposition of calcium carbonate:
Calculate the pressure in atm of CO2 in an equilib-
rium process (a) at 258C and (b) at 8008C. Assume
that DH 8 5 177.8 kJ/mol and DS8 5 160.5 J/K ? mol
for the temperature range.
18.30 The equilibrium constant KP for the reaction
is 5.62 3 1035 at 258C. Calculate DG°f for COCl2 at
258C.
18.31 At 258C, DG8 for the process
is 8.6 kJ/mol. Calculate the vapor pressure of water
at this temperature.
18.32 Calculate DG 8 for the process
Is the formation of graphite from diamond favored at
258C? If so, why is it that diamonds do not become
graphite on standing?
Thermodynamics in Living Systems
Review Questions
18.33 What is a coupled reaction? What is its importance
in biological reactions?
18.34 What is the role of ATP in biological reactions?
Problems
18.35 Referring to the metabolic process involving glucose
on p. 633, calculate the maximum number of moles
C(diamond) ¡ C(graphite)
H2O(l) ∆ H2O(g)
CO(g) 1 Cl2(g) ∆ COCl2(g)
CaCO3(s) ∆ CaO(s) 1 CO2(g)
PCO 5 1.20 atmPH2O 5 0.66 atm
PCO25 0.78 atmPH2
5 0.25 atm,
H2(g) 1 CO2(g) ∆ H2O(g) 1 CO(g)
Cl2PCl3
PCl5(g) ∆ PCl3(g) 1 Cl2(g)
2H2O(g) ∆ 2H2(g) 1 O2(g)
of ATP that can be synthesized from ADP from the
breakdown of one mole of glucose.
18.36 In the metabolism of glucose, the first step is the
conversion of glucose to glucose 6-phosphate:
Because DG8 is positive, this reaction does not favor
the formation of products. Show how this reaction
can be made to proceed by coupling it with the hy-
drolysis of ATP. Write an equation for the coupled
reaction and estimate the equilibrium constant for
the coupled process.
Additional Problems
18.37 Explain the following nursery rhyme in terms of the
second law of thermodynamics.
Humpty Dumpty sat on a wall;
Humpty Dumpty had a great fall.
All the King’s horses and all the King’s men
Couldn’t put Humpty together again.
18.38 Calculate DG for the reaction
at 258C for the following conditions:
(a) [H1] 5 1.0 3 1027 M, [OH2] 5 1.0 3 1027 M
(b) [H1] 5 1.0 3 1023 M, [OH2] 5 1.0 3 1024 M
(c) [H1] 5 1.0 3 10212 M, [OH2] 5 2.0 3 1028 M
(d)
18.39 Which of the following thermodynamic functions
are associated only with the first law of thermody-
namics: S, E, G, and H?
18.40 A student placed 1 g of each of three compounds A,
B, and C in a container and found that after 1 week
no change had occurred. Offer some possible expla-
nations for the fact that no reactions took place. As-
sume that A, B, and C are totally miscible liquids.
18.41 Give a detailed example of each of the following, with
an explanation: (a) a thermodynamically spontaneous
process; (b) a process that would violate the first law
of thermodynamics; (c) a process that would violate
the second law of thermodynamics; (d) an irreversible
process; (e) an equilibrium process.
18.42 Predict the signs of DH, DS, and DG of the system
for the following processes at 1 atm: (a) ammonia
melts at 2608C, (b) ammonia melts at 277.78C, (c)
ammonia melts at 21008C. (The normal melting
point of ammonia is 277.78C.)
18.43 Consider the following facts: Water freezes sponta-
neously at 258C and 1 atm, and ice has a more ordered
structure than liquid water. Explain how a spontaneous
process can lead to a decrease in entropy.
[H1] 5 3.5 M, [OH2] 5 4.8 3 1024 M
H2O(l) ∆ H1(aq) 1 OH2(aq)
¢G° 5 13.4 kJ/mol
glucose 1H3PO4 ¡ glucose 6-phosphate 1 H2O
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638 CHAPTER 18 Thermodynamics
18.44 Ammonium nitrate (NH4NO3) dissolves sponta-
neously and endothermically in water. What can you
deduce about the sign of DS for the solution process?
18.45 Calculate the equilibrium pressure of CO2 due to the
decomposition of barium carbonate (BaCO3) at
258C.
18.46 (a) Trouton’s rule states that the ratio of the molar
heat of vaporization of a liquid (DHvap) to its boiling
point in kelvins is approximately 90 J/K ? mol. Use
the following data to show that this is the case and
explain why Trouton’s rule holds true:
tbp(8C) DHvap(kJ/mol)
Benzene 80.1 31.0
Hexane 68.7 30.8
Mercury 357 59.0
Toluene 110.6 35.2
(b) Use the values in Table 12.5 to calculate the same
ratio for ethanol and water. Explain why Trouton’s
rule does not apply to these two substances as well
as it does to other liquids.
18.47 Referring to Problem 18.46, explain why the ratio is
considerably smaller than 90 J/K ? mol for liquid HF.
18.48 Carbon monoxide (CO) and nitric oxide (NO) are pol-
luting gases contained in automobile exhaust. Under
suitable conditions, these gases can be made to react to
form nitrogen (N2) and the less harmful carbon diox-
ide (CO2). (a) Write an equation for this reaction.
(b) Identify the oxidizing and reducing agents. (c) Cal-
culate the KP for the reaction at 258C. (d) Under nor-
mal atmospheric conditions, the partial pressures are
PN25 0.80 atm, PCO2
5 3.0 3 1024 atm, PCO 5
5.0 3 1025 atm, and PNO 5 5.0 3 1027 atm. Calcu-
late QP and predict the direction toward which the
reaction will proceed. (e) Will raising the tempera-
ture favor the formation of N2 and CO2?
18.49 For reactions carried out under standard-state condi-
tions, Equation (18.10) takes the form DG8 5 DH8 2
TDS8. (a) Assuming DH 8 and DS 8 are independent
of temperature, derive the equation
where K1 and K2 are the equilibrium constants at
T1 and T2, respectively. (b) Given that at 258C Kc is
4.63 3 1023 for the reaction
calculate the equilibrium constant at 658C.
18.50 The Ksp of AgCl is given in Table 17.2. What is its
value at 608C? [Hint: You need the result of Prob-
lem 18.49(a) and the data in Appendix 2 to calcu-
late DH 8.]
N2O4(g) ∆ 2NO2(g) ¢H° 5 58.0 kJ/mol
ln K2
K1
5¢H°
RaT2 2 T1
T1T2
b
18.51 Under what conditions does a substance have a stan-
dard entropy of zero? Can a substance ever have a
negative standard entropy?
18.52 Water gas, a mixture of H2 and CO, is a fuel made by
reacting steam with red-hot coke (a by-product of
coal distillation):
From the data in Appendix 2, estimate the tempera-
ture at which the reaction begins to favor the forma-
tion of products.
18.53 Consider the following Brønstead acid-base reaction
at 258C:
(a) Predict whether K will be greater or smaller than
unity. (b) Does DS8 or DH8 make a greater contribution
to DG8? (c) Is DH8 likely to be positive or negative?
18.54 Crystallization of sodium acetate from a supersatu-
rated solution occurs spontaneously (see p. 426).
What can you deduce about the signs of DS and DH?
18.55 Consider the thermal decomposition of CaCO3:
The equilibrium vapor pressures of CO2 are
22.6 mmHg at 7008C and 1829 mmHg at 9508C.
Calculate the standard enthalpy of the reaction.
[Hint: See Problem 18.49(a).]
18.56 A certain reaction is spontaneous at 728C. If the en-
thalpy change for the reaction is 19 kJ/mol, what is the
minimum value of DS (in (J/K ? mol) for the reaction?
18.57 Predict whether the entropy change is positive or
negative for each of these reactions:
(a)
(b)
(c)
(d)
18.58 The reaction pro-NH3(g) 1 HCl(g) ¡ NH4Cl(s)
2H2O2(l) ¡ 2H2O(l) 1 O2(g)
NH4NO3(s) ¡ N2O(g) 1 2H2O(g)
O(g) 1 O(g) ¡ O2(g)
Zn(s) 1 2HCl(aq) ¡ ZnCl2(aq) 1 H2(g)
CaCO3(s) ∆ CaO(s) 1 CO2(g)
HF(aq) 1 Cl2(aq) ∆ HCl(aq) 1 F2(aq)
H2O(g) 1 C(s) ∆ CO(g) 1 H2(g)
ceeds spontaneously at 258C even though there is a de-
crease in the number of microstates of the system
(gases are converted to a solid). Explain.
18.59 Use the following data to determine the normal boil-
ing point, in kelvins, of mercury. What assumptions
must you make in order to do the calculation?
Hg(l): DH°f 5 0 (by definition)
S° 5 77.4 J/K ? mol
Hg(g): DH°f 5 60.78 kJ/mol
S° 5 174.7 J/K ? mol
18.60 The molar heat of vaporization of ethanol is
39.3 kJ/mol and the boiling point of ethanol is 78.38C.
Calculate DS for the vaporization of 0.50 mol ethanol.
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Questions and Problems 639
18.61 A certain reaction is known to have a DG 8 value of
2122 kJ/mol. Will the reaction necessarily occur if
the reactants are mixed together?
18.62 In the Mond process for the purification of nickel,
carbon monoxide is reacted with heated nickel to
produce Ni(CO)4, which is a gas and can therefore
be separated from solid impurities:
Given that the standard free energies of formation of
CO(g) and Ni(CO)4(g) are 2137.3 kJ/mol and
2587.4 kJ/mol, respectively, calculate the equilib-
rium constant of the reaction at 808C. Assume that
DG°f is temperature independent.
18.63 Calculate DG 8 and KP for the following processes at
258C:
(a)
(b)
Account for the differences in DG 8 and KP obtained
for (a) and (b).
18.64 Calculate the pressure of O2 (in atm) over a sample
of NiO at 258C if DG 8 5 212 kJ/mol for the reaction
18.65 Comment on the statement: “Just talking about en-
tropy increases its value in the universe.”
18.66 For a reaction with a negative DG 8 value, which of
the following statements is false? (a) The equilib-
rium constant K is greater than one, (b) the reaction
is spontaneous when all the reactants and products
are in their standard states, and (c) the reaction is al-
ways exothermic.
18.67 Consider the reaction
Given that DG 8 for the reaction at 258C is 173.4
kJ/mol, (a) calculate the standard free energy of for-
mation of NO, and (b) calculate KP of the reaction.
(c) One of the starting substances in smog formation
is NO. Assuming that the temperature in a running
automobile engine is 11008C, estimate KP for the
above reaction. (d) As farmers know, lightning helps
to produce a better crop. Why?
18.68 Heating copper(II) oxide at 4008C does not produce
any appreciable amount of Cu:
CuO(s) 34 Cu(s) 1 O2(g) DG° 5 127.2 kJ/mol
However, if this reaction is coupled to the conversion of
graphite to carbon monoxide, it becomes spontaneous.
Write an equation for the coupled process and calculate
the equilibrium constant for the coupled reaction.
18.69 The internal combustion engine of a 1200-kg car is de-
signed to run on octane (C8H18), whose enthalpy of
combustion is 5510 kJ/mol. If the car is moving up a
slope, calculate the maximum height (in meters) to
12
N2(g) 1 O2(g) ∆ 2NO(g)
NiO(s) ∆ Ni(s) 112O2(g)
12H2(g) 1
12Br2(l) ∆ HBr(g)
H2(g) 1 Br2(l) ∆ 2HBr(g)
Ni(s) 1 4CO(g) ∆ Ni(CO)4(g)
which the car can be driven on 1.0 gallon of the fuel.
Assume that the engine cylinder temperature is 22008C
and the exit temperature is 7608C, and neglect all forms
of friction. The mass of 1 gallon of fuel is 3.1 kg. [Hint:
The efficiency of the internal combustion engine, de-
fined as work performed by the engine divided by the
energy input, is given by (T2 2 T1)/T2, where T2 and T1
are the engine’s operating temperature and exit tem-
perature (in kelvins). The work done in moving the
car over a vertical distance is mgh, where m is the
mass of the car in kg, g the acceleration due to gravity
(9.81 m/s2), and h the height in meters.]
18.70 A carbon monoxide (CO) crystal is found to have en-
tropy greater than zero at absolute zero of temperature.
Give two possible explanations for this observation.
18.71 (a) Over the years there have been numerous claims
about “perpetual motion machines,” machines that
will produce useful work with no input of energy.
Explain why the first law of thermodynamics pro-
hibits the possibility of such a machine existing. (b)
Another kind of machine, sometimes called a
“perpetual motion of the second kind,” operates as
follows. Suppose an ocean liner sails by scooping up
water from the ocean and then extracting heat from
the water, converting the heat to electric power to
run the ship, and dumping the water back into the
ocean. This process does not violate the first law of
thermodynamics, for no energy is created—energy
from the ocean is just converted to electrical energy.
Show that the second law of thermodynamics pro-
hibits the existence of such a machine.
18.72 The activity series in Section 4.4 shows that reaction
(a) is spontaneous while reaction (b) is nonsponta-
neous at 258C:
(a)
(b)
Use the data in Appendix 2 to calculate the equilib-
rium constant for these reactions and hence confirm
that the activity series is correct.
18.73 The rate constant for the elementary reaction
is 7.1 3 109/M2? s at 258C. What is the rate constant
for the reverse reaction at the same temperature?
18.74 The following reaction was described as the cause of
sulfur deposits formed at volcanic sites:
It may also be used to remove SO2 from powerplant
stack gases. (a) Identify the type of redox reaction it
is. (b) Calculate the equilibrium constant (KP) at 258C
and comment on whether this method is feasible for
removing SO2. (c) Would this procedure become
more or less effective at a higher temperature?
2H2S(g) 1 SO2(g) ∆ 3S(s) 1 2H2O(g)
2NO(g) 1 O2(g) ¡ 2NO2(g)
Cu(s) 1 2H1¡ Cu21(aq) 1 H2(g)
Fe(s) 1 2H1¡ Fe21(aq) 1 H2(g)
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640 CHAPTER 18 Thermodynamics
18.75 Describe two ways that you could measure DG 8 of a
reaction.
18.76 The following reaction represents the removal of
ozone in the stratosphere:
Calculate the equilibrium constant (KP) for the reac-
tion. In view of the magnitude of the equilibrium con-
stant, explain why this reaction is not considered a
major cause of ozone depletion in the absence of man-
made pollutants such as the nitrogen oxides and
CFCs? Assume the temperature of the stratosphere to
be 2308C and DG8f to be temperature independent.
18.77 A 74.6-g ice cube floats in the Arctic Sea. The tem-
perature and pressure of the system and surround-
ings are at 1 atm and 08C. Calculate DSsys, DSsurr,
and DSuniv for the melting of the ice cube. What can
you conclude about the nature of the process from
the value of DSuniv? (The molar heat of fusion of
water is 6.01 kJ/mol.)
18.78 Comment on the feasibility of extracting copper
from its ore chalcocite (Cu2S) by heating:
Calculate the DG8 for the overall reaction if the above
process is coupled to the conversion of sulfur to sulfur
dioxide, given that DG°f (Cu2S) 5 286.1 kJ/mol.
18.79 Active transport is the process in which a substance is
transferred from a region of lower concentration to one
of higher concentration. This is a nonspontaneous
process and must be coupled to a spontaneous process,
such as the hydrolysis of ATP. The concentrations
of K1 ions in the blood plasma and in nerve cells
are 15 mM and 400 mM, respectively (1 mM 5
1 3 1023 M). Use Equation (18.13) to calculate DG for
the process at the physiological temperature of 378C:
In this calculation, the DG8 term can be set to zero.
What is the justification for this step?
18.80 Large quantities of hydrogen are needed for the
synthesis of ammonia. One preparation of hydro-
gen involves the reaction between carbon monox-
ide and steam at 3008C in the presence of a
copper-zinc catalyst:
Calculate the equilibrium constant (KP) for the reac-
tion and the temperature at which the reaction favors
the formation of CO and H2O. Will a larger KP be at-
tained at the same temperature if a more efficient
catalyst is used?
18.81 Consider two carboxylic acids (acids that contain
the OCOOH group): CH3COOH (acetic acid,
Ka 5 1.8 3 1025) and CH2ClCOOH (chloroacetic
CO(g) 1 H2O(g) ∆ CO2(g) 1 H2(g)
K1(15 mM) ¡ K1(400 mM)
Cu2S(s) ¡ 2Cu(s) 1 S(s)
2O3(g) ∆ 3O2(g)
acid, Ka 5 1.4 3 1023). (a) Calculate DG8 for the
ionization of these acids at 258C (b) From the equa-
tion DG 8 5 DH 8 2 TDS8, we see that the contribu-
tions to the DG 8 term are an enthalpy term (DH8)
and a temperature times entropy term (TDS 8). These
contributions are listed below for the two acids:
DH8(kJ/mol) TDS8(kJ/mol)
CH3COOH 20.57 227.6
CH2ClCOOH 24.7 221.1
Which is the dominant term in determining the value
of DG8 (and hence Ka of the acid)? (c) What processes
contribute to DH8? (Consider the ionization of the
acids as a Brønsted acid-base reaction.) (d) Explain
why the TDS8 term is more negative for CH3COOH.
18.82 One of the steps in the extraction of iron from its ore
(FeO) is the reduction of iron(II) oxide by carbon
monoxide at 9008C:
If CO is allowed to react with an excess of FeO, cal-
culate the mole fractions of CO and CO2 at equilib-
rium. State any assumptions.
18.83 Derive the following equation
where Q is the reaction quotient and describe how you
would use it to predict the spontaneity of a reaction.
18.84 The sublimation of carbon dioxide at 2788C is given
by
Calculate DSsub when 84.8 g of CO2 sublimes at this
temperature.
18.85 Entropy has sometimes been described as “time’s ar-
row” because it is the property that determines the
forward direction of time. Explain.
18.86 Referring to Figure 18.1, we see that the probability
of finding all 100 molecules in the same flask is 8 3
10231 (see Problem 18.6). Assuming that the age of
the universe is 13 billion years, calculate the time in
seconds during which this event can be observed.
18.87 A student looked up the DG°f, DH 8f, and S8 values for
CO2 in Appendix 2. Plugging these values into
Equation (18.10), he found that DG°f Þ DH 8f 2 TS 8
at 298 K. What is wrong with his approach?
18.88 Consider the following reaction at 298 K:
Calculate DSsys, DSsurr, and DSuniv for the reaction.
18.89 As an approximation, we can assume that proteins
exist either in the native (or physiologically func-
tioning) state and the denatured state
¢H° 5 2571.6 kJ/mol
2H2(g) 1 O2(g) ¡ 2H2O(l)
¢Hsub 5 25.2 kJ/molCO2(s) ¡ CO2(g)
¢G 5 RT ln (Q/K)
FeO(s) 1 CO(g) ∆ Fe(s) 1 CO2(g)
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Answers to Practice Exercises 641
The standard molar enthalpy and entropy of the de-
naturation of a certain protein are 512 kJ/mol and
1.60 kJ/K ? mol, respectively. Comment on the signs
and magnitudes of these quantities, and calculate the
temperature at which the process favors the dena-
tured state.
18.90 Which of the following are not state functions: S, H,
q, w, T?
18.91 Which of the following is not accompanied by an in-
crease in the entropy of the system? (a) mixing of
native ∆ denatured two gases at the same temperature and pressure,
(b) mixing of ethanol and water, (c) discharging a
battery, (d) expansion of a gas followed by compres-
sion to its original temperature, pressure, and volume.
18.92 Hydrogenation reactions (for example, the process
of converting CPC bonds to COC bonds in food
industry) are facilitated by the use of a transition
metal catalyst, such as Ni or Pt. The initial step is the
adsorption, or binding, of hydrogen gas onto the
metal surface. Predict the signs of DH, DS, and DG
when hydrogen gas is adsorbed onto the surface of
Ni metal.
18.93 The standard free energies of formation (DG°f) of the
three isomers of pentane (see p. 358) in the gas phase
are: n-pentane: 28.37 kJ/mol; 2-methylbutane: 214.8
kJ/mol; 2,2-dimethylpropane: 215.2 kJ/mol. (a) Deter-
mine the mole percent of these molecules in an equilib-
rium mixture at 258C. (b) How does the stability of
these molecules depend on the extent of branching?
18.94 Carry out the following experiments: Quickly stretch a
rubber band (at least 0.5 cm wide) and press it against
your lips. You will feel a warming effect. Next, carry
out the reverse procedure; that is, first stretch a rubber
band and hold it in position for a few seconds. Now
quickly release the tension and press the rubber band
against your lips. You will feel a cooling effect. (a) Ap-
ply Equation (18.7) to these processes to determine the
signs of DG, DH, and hence DS in each case. (b) From
the signs of DS, what can you conclude about the
structure of rubber molecules?
18.95 At 0 K, the entropy of carbon monoxide crystal is
not zero but has a value of 4.2 J/K ? mol, called the
residual entropy. According to the third law of ther-
modynamics, this means that the crystal does not
have a perfect arrangement of the CO molecules.
(a) What would be the residual entropy if the
arrangement were totally random? (b) Comment on
the difference between the result in (a) and
4.2 J/K ? mol [Hint: Assume that each CO molecule
has two choices for orientation and use Equation
(18.1) to calculate the residual entropy.]
18.96 Comment on the correctness of the analogy some-
times used to relate a student’s dormitory room be-
coming untidy to an increase in entropy.
18.97 The standard enthalpy of formation and the standard
entropy of gaseous benzene are 82.93 kJ/mol and
269.2 J/K ? mol, respectively. Calculate DH 8, DS 8,
and DG 8 for the process at 258C. Comment on your
answers.
18.98 The following diagram shows the variation of the equi-
librium constant with temperature for the reaction
Calculate DG 8, DH 8, and DS 8 for the reaction at
872 K. (Hint: See Problem 18.49.)
I2(g) ∆ 2I(g)
C6H6(l) ¡ C6H6(g)
SPECIAL PROBLEMS
ln K
T2 5 1173 K T1 5 872 K
K1 5 1.80 3 1024
K2 5 0.0480
1DT
ANSWERS TO PRACTICE EXERCISES
18.1 (a) Entropy decreases, (b) entropy decreases,
(c) entropy increases, (d) entropy increases.
18.2 (a) 2173.6 J/K ? mol, (b) 2139.8 J/K ? mol, (c) 215.3
J/K ? mol. 18.3 (a) DS . 0, (b) DS , 0, (c) DS ø 0.
18.4 (a) 2106.4 kJ/mol, (b) 22935.0 kJ/mol.
18.5 DSfus 5 16 J/K ? mol; DSvap 5 72 J/K ? mol.
18.6 2 3 1057. 18.7 33 kJ/mol. 18.8 DG 5 0.97
kJ/mol; direction is from right to left.
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